Question

Evaluate the integral.

\(\int_0^1 {\cosh } tdt\).

Short answer

The solution of given integral\(\int_0^1 {\cosh } tdt\) is \(1.175\).

Step-by-step solution

Step 1:Definition of integration or Antiderivative

The function\({\rm{F}}\)is called as antiderivative of\({\rm{f}}\)on an interval\({\rm{I}}\)if\({\rm{F'(x) = f(x)}}\)

i.e.,\(\int {{\rm{f(x)dx = F(x)}}} \).

Applying definition of integration

Using evaluation theorem writes given integral as-

\(\begin{aligned}{c}\int_0^1 {\cosh } (t)dt = \left. {\sinh (t)} \right|_0^1\\ = \frac{{{e^1} - {e^{ - 1}}}}{2} - \frac{{{e^0} - {e^{ - 0}}}}{2}\\ = \frac{{{e^1} - {e^{ - 1}}}}{2}\\ \approx 1.175\end{aligned}\)

Therefore, the solution of given integral\(\int_0^1 {\cosh } tdt\) is \(1.175\).